Which of the following ions has the maximum m | vedclass

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(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.
For $

Vedclass · Accepted Answer

$Fe^{3+}$

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(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.
For $V^{3+}$ $([Ar] 3d^2)$: $n = 2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 	ext{ BM}$.
For $Mn^{3+}$ $([Ar] 3d^4)$: $n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 	ext{ BM}$.
For $Fe^{3+}$ $([Ar] 3d^5)$: $n = 5$,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 	ext{ BM}$.
For $Cu^{2+}$ $([Ar] 3d^9)$: $n = 1$,$\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73 	ext{ BM}$.
Thus,$Fe^{3+}$ has the maximum number of unpaired electrons $(n=5)$,resulting in the maximum magnetic moment.

Which of the following ions has the maximum magnetic moment?

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